Question

The oxidation of ferrous oxalate by \(KMnO_4\) in acidic medium is given by the following equation, \(3MnO_4^- + 5FeC_2O_4 + 24H^+ \to 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O\). What is the volume of \(0.01 \text{ mol dm}^{-3}\) \(KMnO_4\) required to oxidise \(15 \text{ cm}^3\) of an acidified solution of \(0.01 \text{ mol dm}^{-3}\) ferrous oxalate?

• A. \(18 \text{ cm}^3\)
• B. \(9 \text{ cm}^3\)
• C. \(15 \text{ cm}^3\)
• D. \(18 \text{ cm}^3\)

Hint:

In acidic medium, \(KMnO_4\) (\(Mn^{+7} \to Mn^{+2}\)) has an n-factor of 5. For \(FeC_2O_4\), \(Fe^{2+} \to Fe^{3+}\) (1 electron) and \(C_2O_4^{2-} \to 2CO_2\) (2 electrons), giving a total n-factor of 3. Use \(N_1V_1 = N_2V_2\) where \(N = M \times \text{n-factor}\).
\((0.01 \times 5) \times V_1 = (0.01 \times 3) \times 15 \implies V_1 = (3 \times 15) / 5 = 9\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In this redox titration, we use the stoichiometry of the balanced chemical equation to find the required volume of the titrant (\(KMnO_4\)). The mole ratio between the oxidant and the reductant is key.

Step 2: Key Formula or Approach:
From the balanced equation:
Moles of \(MnO_4^-\) Moles of \(FeC_2O_4 = 3 / 5\).
Molar concentration (\(M\)) \(\times\) Volume (\(V\)) = Moles (\(n\)).

Step 3: Detailed Explanation:
1. Calculate the moles of ferrous oxalate (\(FeC_2O_4\)):
\[ n_{oxalate} = M \times V = 0.01 \text{ mol/dm}^3 \times (15/1000) \text{ dm}^3 = 1.5 \times 10^{-4} \text{ mol} \]
2. Use the stoichiometric ratio to find the required moles of \(KMnO_4\):
\[ \frac{n_{KMnO_4}}{n_{oxalate}} = \frac{3}{5} \]
\[ n_{KMnO_4} = \frac{3}{5} \times 1.5 \times 10^{-4} \text{ mol} = 0.9 \times 10^{-4} \text{ mol} \]
3. Calculate the volume of \(KMnO_4\) solution:
\[ V_{KMnO_4} = \frac{n}{M} = \frac{0.9 \times 10^{-4} \text{ mol}}{0.01 \text{ mol/dm}^3} = 9 \times 10^{-3} \text{ dm}^3 \]
Converting to \(cm^3\) (\(1 \text{ dm}^3 = 1000 \text{ cm}^3\)):
\[ V = 9 \text{ cm}^3 \]

Step 4: Final Answer:
The volume of \(KMnO_4\) required is \(9 \text{ cm}^3\).